Physics question deals with circular motion.?

Question

A stone has a mass of 6.0 x10^-3 kg and is wedged into the tread of an automobile tire, as the drawing shows. The coefficient of static friction between the stone and each side of the tread channel is 0.90. When the tire surface is rotating at a maximum speed of 13 m/s, the stone flies out of the tread. The magnitude FN of the normal force that each side of the tread channel exerts on the stone is 1.8 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire.
Well I basically worked till I got this equation r=v^2/usFN
(from r=v^2/usg and v^2=rusFN/m)

so plugging in the numbers I got r=(14.1^2) (6.93x10^-3)/ (.844)(2.44) which r= .289946 but this isn't correct according to the answer key. I really need help on how to work this problem. Thanks in advance.

Answer 1

You are on the right track
a=v^2/r
F=ma=mv^2/r
F=f=uN (coefficient of friction times the normal force)

so
r=mv^2/(uN)
r=6.0 e-3 (13 )^2/(.9 x 1.8)=0.626m
or
r=62.6 cm

I do not know what you have done and why, but this is my train of thought.
I hope it was useful

Answer 2

simply by fact the disc rotates a million revolution, the factor on the rim strikes a distance of the circumference of the circle. Circumference = 2 * ? * r Radius ½ * 23.0 = 11.5 cm = 0.one hundred fifteen m Circumference = 2 * ? * 0.one hundred fifteen m sixteen.6 revolutions = sixteen.6 * 2 * ? * 0.one hundred fifteen it relatively is the gap that the factor strikes in a million minute velocity = distance that the factor strikes in a million 2nd velocity = sixteen.6 * 2 * ? * 0.one hundred fifteen ÷ 60 = 0.2 m/s it relatively is the linear velocity of a factor on the rim. The centripetal acceleration = v^2/r = 0.2^2/0.one hundred fifteen = 0.348 m/s^2